A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is P × 10¹¹ Nm⁻², where the value of P is: (Take g = 3π m/s²)

Solution

Physics sir · 4 steps

Write Young's modulus as FL/(A Δ L) and watch the π in g cancel the π in the area.

1. The formula. Young's modulus is stress over strain. Stress is F/A and strain is Δ L/L, so dividing one by the other gives the expression below.

   Y = (F/A)/(Δ L/L) = mgL/(π r² Δ L)

2. Convert to SI. The load is F = mg = 50 × 3π = 150π N. The area is A = π r² = π (3 × 10⁻³)² = 9π × 10⁻⁶ m². The extension is Δ L = 0.1 mm = 10⁻⁴ m.
3. Put in numbers. Y = (150π × 3)/(9π × 10⁻⁶ × 10⁻⁴). The π cancels, which leaves 450/(9 × 10⁻¹⁰) = 50 × 10¹⁰ = 5 × 10¹¹ N m⁻².
4. So. Y = 5 × 10¹¹ N m⁻², so P = 5, which is option (d).

The idea to keep. For a loaded wire, Y = mgL/(π r² Δ L). Convert every millimetre to metres before you substitute. An odd value of g such as 3π is usually there so that the π cancels.

Answer from the NTA final answer key. Solution by PrepWiser.